UESTION 5 If \( \cos 21^{\circ}=p \) determine ile following in terms of \( p \). \( 5.1 .1 \quad \tan 201^{\circ} \) \( 5.1 .2 \sin 42^{\circ} \) (3) \( 5.1 .3 \quad \cos 51^{\circ} \) 2 Simplify \( : \frac{\sin 210^{\circ} \cdot \cos 510^{\circ}}{\cos 315^{\circ} \cdot \sin \left(-135^{\circ}\right)} \) (3) 3 Prove the identity: \[ \frac{\cos \theta-\cos 2 \theta+2}{3 \sin \theta-\sin 2 \theta}=\frac{1+\cos \theta}{\sin \theta} \] 4 Determine the general solution of \( \sin \theta \sin \frac{30}{2}+\cos \frac{3 \theta}{2} \cos \theta=-\frac{\sqrt{3}}{2} \). Given: \( \sin \theta \cdot \cos \beta=-1 \) 5.5.1 Write down the maximum and ninimum value of \( \cos \beta \) 5.5.2 Solve for \( \theta=\left[0^{\circ} ; 270^{\circ}\right] \) and \( \beta \in\left[-180^{\circ}, 70^{\circ}\right] \).
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To find \( \tan 201^{\circ} \) in terms of \( p \), we can use the identity \( \tan(180^{\circ} + x) = \tan x \). So, \( \tan 201^{\circ} = \tan(21^{\circ}) = \frac{\sin 21^{\circ}}{\cos 21^{\circ}} = \frac{\sqrt{1 - p^{2}}}{p} \), where \( p = \cos 21^{\circ} \). Next up, for \( \sin 42^{\circ} \), we can utilize the double angle identity, \( \sin 2\theta = 2 \sin \theta \cos \theta \). Therefore, \( \sin 42^{\circ} = \sin(2 \times 21^{\circ}) = 2 \sin 21^{\circ} \cos 21^{\circ} = 2 \sqrt{1 - p^{2}} \cdot p \).
